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Is $\phi$ set forms a metric space or not ?

I think, it does not form a metric space, because, we can't specify a metric on $\phi$.

But, In many text book, it is not mention that, the set on which, we define metric should be non empty.

If I may suppose, that d is a function define on $\phi$ $ \times$ $\phi$ such that

d is constant function with range set { $0$ }. Then it must be metric on {$\phi$}.

Plz help... what is the right thingh ?

ram
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    By {$\phi$}, do you mean the empty set? – Chris Eagle Nov 02 '12 at 14:19
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    First, $\phi$ is not the letter for the empty set. $\varnothing,\emptyset$ are the notation for the empty set. Secondly note that ${\varnothing}$ is the set whose only element is the empty set. In particular ${\varnothing}$ is not empty, and therefore ${\varnothing}\neq\varnothing$. – Asaf Karagila Nov 02 '12 at 14:23
  • You can specify whatever metric you want as it is vacuous. – copper.hat Nov 02 '12 at 14:24
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    Well, you can't specify whatever metric you want. The metric would have to be a function $d \colon \varnothing \times \varnothing \to \mathbb{R}$. But $\varnothing \times \varnothing = \varnothing$, and there is a unique function from the empty set to any other set. Thus $d \colon \varnothing \to \mathbb{R}$ must be the "empty function." – Manny Reyes Nov 02 '12 at 14:47
  • With all due respect, why do you worry about this actually? – Hui Yu Nov 02 '12 at 15:38
  • Why are metric spaces non-empty? might be related (you still haven't specified whether you mean the empty space or the space containing the empty set as only point). – Martin Jan 03 '13 at 07:35

2 Answers2

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Yes. A singleton is a metric space.

To see that, note that a subspace of a metric space is a metric space, and $\{x\}$ is a subset of $\mathbb R$ whenever $x\in\mathbb R$.

Since all singletons are "essentially" the same, this means that $\{\varnothing\}$ can also be thought as a metric space.

On the other hand whether or not $\varnothing$ itself, the empty set, is a metric space is up to definition, whether or not you are allowing empty structures in your universe, or does the empty set carries no structure.

Asaf Karagila
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Obviusly $(\{\emptyset\},d)$ defines a metric space, the trivial one. Or you can say vacuously. You can check that the rules of a metric space are satisfied.

Michael Albanese
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ILoveMath
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